question_answer

A metallic rod of length \[l\] is placed normal to the magnetic field B and revolved in a circular path about one of the ends with angular frequency m. The potential difference across the ends will be:

A)  \[\frac{1}{2}{{B}^{2}}l\omega \]

B)                                        \[\frac{1}{2}B\omega {{l}^{2}}\]

C)         \[\frac{1}{8}B\omega {{l}^{3}}\]                              

D)         \[B\omega {{l}^{3}}\]

Correct Answer: B

Solution :

Suppose a conducting rod of length \[l\]rotates with a constant angular speed \[\omega \] about a point at one end. A uniform magnetic field \[\vec{B}\]is directed perpendicular to the plane of rotation as shown in figure. Consider a segment of rod of length drat a distance r from O. This segment has a velocity \[v=r\omega .\] The induced emf in this segment is \[de=Bvdr=B(r\omega )dr\] Summing the emfs induced across all segments, which are in series, gives the total emf across the rod.                                                     \[\therefore \]  \[e=\int_{0}^{1}{de}=\int_{0}^{1}{Br\omega dr}=\frac{B\omega {{l}^{2}}}{2}\]                 \[\therefore \]  \[e=\frac{B\omega {{l}^{2}}}{2}\] rom right hand rule we can see that P is at higher potential than O. Thus,     \[{{V}_{P}}-{{V}_{O}}=\frac{B\omega {{l}^{2}}}{2}\]